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Riddles and Answers © 2024

Displayed In December

Hint:
A nativity scene
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Under The Cup Riddle

Hint: Write down the possibilities. Remember that there are only three cups, so if the rightmost cup wasn't touched...
The rightmost cup.

The rightmost cup has a half chance of holding the coin, and the other cups have a quarter chance.

Pretend that Os represent cups, and Q represents the cup with the coin.

The game starts like this:

OOQ

Then your friend switches the rightmost cup with another, giving two possibilities, with equal chance:

OQO
QOO

Your friend then moves the cups again, but doesn't touch the rightmost cup. The only switch possible is with the leftmost cup and the middle cup. This gives two possibilities with equal chance:

QOO
OQO

Lastly, your friend switches the rightmost cup with another cup. If the first possibility shown above was true, there would be two possibilities, with equal chance:

OOQ
QOO

If the second possibility shown above (In the second switch) was true, there would be two possibilities with equal chance:

OOQ
OQO

This means there are four possibilities altogether, with equal chance:

OOQ
QOO
OOQ
OQO

This means each possibility equals to a quarter chance, and because there are two possibilities with the rightmost cup having the coin, there is a half chance that the coin is there.
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The Loaded Revolver Riddle

Hint:
Henry should have Gretchen pull the trigger again without spinning.

We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.

If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
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Roll The Dice

Hint: What will happen if there are 6 gamblers, each of whom bet on a different number?
It's a fair game. If there are 6 gamblers, each of whom bet on a different number, the dealer will neither win nor lose on each deal.

If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.

If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.

If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.

In each case, the dealer neither wins nor loses. Hence it's a fair game.
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Knights Of The Round Table Riddle

Hint: Does it matter if they are sitting clockwise or counterclockwise? Or where the oldest sits?
The odds are 11:1. (The probability is 1/12.)

Imagine they sat down in age order, with each person randomly picking a seat. The first person is guaranteed to pick a seat that "works". The second oldest can sit to his right or left, since these five can sit either clockwise or counterclockwise. The probability of picking a seat that works is thus 2/4, or 1/2. The third oldest now has three chairs to choose from, one of which continues the progression in the order determined by the second person, for a probability of 1/3. This leaves two seats for the fourth oldest, or a 1/2 chance. The youngest would thus be guaranteed to sit in the right seat, since there is only one seat left. This gives 1 * 1/2 * 1/3 * 1/2 * 1 = 1/12, or 11:1 odds against.
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Chances Of A 2nd Girl Riddle

Hint: It is not 1/2 as you would first think.
1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Boy - Girl
Girl - Boy
Boy - Boy
Girl - Girl

Since we know one of the children is a boy, we will drop the girl-girl possibility from the sample space.
This leaves only three possibilities, one of which is two boys. Hence the probability is 1/3
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Hint: Their dad is a very smart person.
Believe it or not, both Mike and James have a 1/2 chance of winning.

James wins if:
-he calls both coin flips right = 1/2 x 1/2 = 1/4
OR
-he does not call both coin flips right, Mike does not call the die roll correctly, and he guesses the number on the spinner right = 3/4 x 5/6 x 2/5 = 30/120 = 1/4

1/4 + 1/4 = 1/2

Mike wins if:
-James does not call both coin flips right and he calls the die roll correctly = 3/4 x 1/6 = 3/24 = 1/8
OR
-James does not call both coin flips right, he does not call the die roll correctly, and Mike does not guess the number on the spinner right = 3/4 x 5/6 x 3/5 = 45/120 = 3/8

1/8 + 3/8 = 1/2

Of course, dad could have just flipped a coin
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The Gardners Riddle

Hint:
Gretchen said that there were 4 girls in the family, three of whom were blond.

This would make the probability that she saw two blonds (3/4) * (2/3), which equals 1/2.

Other numbers would work, but the next pair would lead to a rather large family.
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The Secret Santa Exchange

Hint: It's not as difficult as it seems. It's the number of ways the friends can form a circle divided by the number of ways the names can be drawn out of the hat.
1/10

For a group of n friends, there are n! (n factorial) ways to draw the names out of the hat. Since a circle does not have a beginning and end, choose one person as the beginning and end of the circle. There are now (n-1)! ways to distribute the remaining people around the circle. Thus the probability of forming a single circle is

(n-1)! / n!

Since n! = (n-1)! * n (for n > 1), this can be rewritten as

(n-1)! / (n*(n-1)!)

Factoring out the (n-1)! from the numerator and denominator leaves

1/n

as the probability.
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Russian Roulette Riddle

Hint:
Russian Roulette
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.


Puzzle ID: #17681
Fun: *** (2.59)
Difficulty: ** (2.07)
Category: Probability
Submitted By: JMCLEOD****
Corrected By: cnmne








You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?


Answer
Label the chambers 1 through 6. Chambers 1 through 3 have bullets and chambers 4 through 6 are empty. After you spin the cylinder there are six possible outcomes:

1. Chamber 1 is fired first: Player 1 loses
2. Chamber 2 is fired first: Player 1 loses
3. Chamber 3 is fired first: Player 1 loses
4. Chamber 4 is fired first: Player 2 loses (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)
5. Chamber 5 is fired first: Player 1 loses (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)
6. Chamber 6 is fired first: Player 2 loses (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)

Therefore player 2 has an 4/6 or 2/3 chance of winning.
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A Thought In Your Mind

Hint:
Memory
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Kindness And Cruelty

Hint:
Fate
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A Fathers Murder

Hint:
So the woman would go to his father's funeral and he can get her number this time....98% of people who got this right turned out to be serial killers...
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Exposed To A Disease Riddle

Hint:
The surgeon is the boys mother.
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Man With A Wooden Leg

Hint:
You cant take a picture with a wooden leg. You need a camera (or iPad or cell phone) to take a picture.
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